I don't know the problem involved Backward difference.

Please load the“dataset.mat”file

(a). Develop first derivative function using Backward finite-difference

function [ diff_f ] = bdiff( f, x, h )

x_i = x;
x_ih = x-h;

diff_f = double((subs(f,x_i) - subs(f,x_ih)) / h);

end

(b). Calculate and display first derivative of given “velocity” data with respect to time

b-1. Unit of velocity is m/s

b-2. Set x-axis 0 to 30 and y-axis -5 to 5

b-3. Display should include differentiation of velocity from your function and “acceleration” data

(c). Caltulate the integration of velocity using Simpson's 1/3 rule ( 0sec to 30sec )

clear all
close all
clc

load('dataset.mat')

x = Time;
y = velocity;
y1 = acceleration;
h = Time(2) - Time(1);
N = 1000;

for i=2:N+1
    bdiff_x = bdiff(y(i),x(i),h);    
end

a = Time(1);
b = Time(2);
n = 30;
S = simpson(y,a,b,n);

figure
plot(x(1:N),bdiff_x,'b--','linewidth',1.5)
hold on
grid on
plot(x,y1,'g-','linewidth',1)
plot(x,S,'r--','linewidth',1)
legend('bDiff','Diff','S')
axis([0 30   -5 5])
xlabel('x')
ylabel('y')

I made a and solved B, but I can't even execute it. C couldn't do it. I need a help.

Answer 1

[Jim Riggs]

I am assuming that your question is that you don't understand Simpson's rule. Below is an exerpt from Wikipedia:

Simoson's Rule is a method of approximating the integral of a function. From the figure, above, Simpson'e rule approximates the integral of f(x) (the blue curve) by using a polynomial function shape (the red curve).

The botom equation shows the parameters for this equation. The method uses pairs of intervals (interval a to m, and m to b in the figure) . There must be an even number if intervals, therefore, there will be an odd number of data points.

Let's assume that your velocity data is in vector Y, which is n samples long (n needs to be an odd number). They also need to be evenly spaced. (interval a to m must be the same length as interval m to b on the independent axis).

You will add up approximations for each pair of intervals in Y, like so:

The first approximation consists of a pair of interval defined by a, m, and b; f(a) = Y(1) and f(m) = Y(2), f(b) = Y(3)

The area under the curve is given by the formula S = (h/3) * [ f(a) + 4 f(m) + f(b)], where h = (b-a)/2 , therefore

S1 = (h/3)  * ( Y(1) + 4*Y(2) + Y(3) );  % this is the integral of Y from Y(1) to Y(3)

Now compute the area for the next 2 intervals:

S2 = (h/3) * ( Y(3) + 4*Y(4) + Y(5) );  % this is the integral of Y from Y(3) to Y(5)

The total area from Y(1) to Y(5) is: S = S1 + S2.

To generalize, you sum up the areas using interval pairs:

h = Time(2) - Time(1);  % independent axis spacing (i.e. the time step)
n = length(Y);  % n must be an odd number
S = 0;
for i=1:2:n  % i will increment by 2 each pass through the loop
    S = S + (h/3) * (Y(i) + 4*Y(i+1) + Y(i+2) );
end

The first pass through this loop, i=1 and the area increment = (h/3)*( Y(1) + 4*Y(2) + Y(3))

The second pass, i increments by 2 and is equal to 3, so the second increment is (h/3)*( Y(3) + 4*Y(4) + Y(5) ).

When the loop completes, S is the area under curveY from Y(1) to Y(n) (which is the distance traveled from Time(1) to Time(n).